# Thread: Simple puzzles to stump people

1. Originally Posted by athenian200
I'm probably wrong, but I have no idea how to proceed. Sorry.
Wait a moment... Isn't non a quadruple amputee?
(You have to be sly and sneaky to solve these sorts of problems!)

2. Originally Posted by Economica
I refer you to post 19 in this thread. Familiarity with Bayes' Rule is not enough.
Familiarity, and familiarity of use are two different things. No offense but learning something in a school context is very different from having a technique in your repertoire that you apply to work, and perhaps seeps in to everyday thinking. When checking for rare diseases, people are always cautioned that a positive is not nearly as certain as people think. People in juries are also told to not take DNA testing accuracy as far as it does, since the particular crime itself could be a rare event.

The 12% answer is an example of using an intermediate number as the answer it self. Were 17% and 29% also common incorrect answers?

Originally Posted by Economica
I refer you also to the Ph.D.'s in math who embarassed themselves telling off Marilyn vos Savant regarding the Monty Hall problem.
I found that funny also. But I think that has to do with the style of probability education received by those PHDs. Some times people just learn mechanisms without understanding the notions behind them. If people were taught the Long-Run Frequency interpretation of probability, then they would automatically prune the tree to situations they needed, instead of attempting a complcated and error prone application of Bayes' Rule.

Originally Posted by Economica
Edit: Oh, and the card problem I got right immediately. (The cab problem, no, obviously. :blushing
But still, I am always amazed when people I think would generally get these types of problems, miss them. So, I probably just shot down my own theory.

Not sure. Maybe people need a little Conceptual Blockbusting, in addition to specific training.

3. I tried to solve the cab problem without looking from anywhere.

Assuming that cab's color does not change it's probability of being seen in a hit-and-run, we have probabilities of cab sightings for the witness:

Blue seen as blue (BB): .15*.8 = .13
Blue seen as green (BG): .15*.2 = .3
Green seen as blue (GB): .85*.2 = .17
Green seen as green (GG): .85*.8 = .68

Chance for a blue sighting to be correct =
Correct sightings of blue / all sightings of blue =
BB / (BB+GB) = .13/(.13 + .17) = .433

oh, I checked the answer. Lol, where did I get that .13???
The correct answer is given with .13 replaced with .12 ..
And I got it wrong!

4. Originally Posted by Jennifer
Wait a moment... Isn't non a quadruple amputee?
(You have to be sly and sneaky to solve these sorts of problems!)
She is? That would change everything, then...

If my previous reasoning was correct, then that would leave 18 possible people for each of the others, and automatically make Nonpareil's count 0. It still doesn't explain exactly why each of them had different numbers from each other, however. I fail to see that part of it.

I wouldn't have thought to seek out that information. It simply didn't cross my mind.

5. Originally Posted by Jennifer
Wait a moment... Isn't non a quadruple amputee?
(You have to be sly and sneaky to solve these sorts of problems!)
Seriously?! Have I been insensitive again?

Originally Posted by athenian200
There are 22 people, then.

The person asking the question doesn't include themselves, leaving 21 people including Nonpareil.

If no one shook hands with their partner, that would leave 19 possible people for each of the others, because they wouldn't shake hands with themselves. Since Nonpareil is part of the couple doing the inviting, it is likely that she shook hands with all possible guests, while the guests might only have shaken hands with the people they knew or were introduced to. Therefore... 20 people?

I'm probably wrong, but I have no idea how to proceed. Sorry.
I won't say yes or no to guesses.

But I will give a hint:
Try smaller number of couples, till you can actually conclusively think through the answer. Then see if you can extrapolate to 10 couples (or really n couples).

6. Originally Posted by ygolo
Hope I wasn't to presumptuous in using your names in the puzzle.

Congratulations Non and pt.
Not at all and thanks.

Oh, the answer is 0... but I have inside information on that one

(And no, she is a fully functional woman. Except for the whole INTJ thing. *shrug* )

7. Okay....

There are 22 people.

Nonpareil may shake hands with everyone except ptgatsby and herself.

Nonpareil may shake hands with 20 people, presuming she has arms/will.

Each other member may shake hands with 19 people if the asker (ptgatsby) is not included in potential shakers, and 20 if he is.

If Nonpareil is armless/unwilling to socialize, then she may shake hands with 0 people, and each of the others may shake hands with 19 people if ptgatsby is included, and 18 if he is not.

There is not enough information to explain why the individual's numbers diverged from one another.

8. Originally Posted by athenian200
Okay....

There are 22 people.

Nonpareil may shake hands with everyone except ptgatsby and herself.

Nonpareil may shake hands with 20 people, presuming she has arms/will.

Each other member may shake hands with 19 people if the asker (ptgatsby) is not included in potential shakers, and 20 if he is.

If Nonpareil is armless/unwilling to socialize, then she may shake hands with 0 people, and each of the others may shake hands with 19 people if ptgatsby is included, and 18 if he is not.

There is not enough information to explain why the individual's numbers diverged from one another.
Seriously, try it for a much smaller problem. Say they only invite 1 couple, what would the answer be?

9. Originally Posted by ygolo
Seriously, try it for a much smaller problem. Say they only invite 1 couple, what would the answer be?
There are 4 people total.

Nonpareil may shake hands with everyone except herself and ptgatsby.

Nonpareil may shake hands with 2 people.

Each of the others may shake hands with only 1 person (Nonpareil) if ptgatsby is not included as a potential shaker, and 2 people if he is.

I don't see where I'm going wrong.

10. Originally Posted by Jennifer
Wait a moment... Isn't non a quadruple amputee?
(You have to be sly and sneaky to solve these sorts of problems!)
Originally Posted by ygolo
Seriously?! Have I been insensitive again?
No one knew there was a shark in the pool.
It was quite a tragedy.

(Note to self: Never bring a keg to a marine biologist's birthday party.)

As far as one couple goes, if everyone shakes a different number of hands, and no one shakes their partner's hand(s), then the number of hands shaken with one couple present would be 0, 1, or 2 (with Non probably shaking 2). Is that right? <-- spoiler

Originally Posted by athenian200
Each of the others may shake hands with only 1 person (Nonpareil) if ptgatsby is not included as a potential shaker, and 2 people if he is.
Well, part of the stipulation is that everyone shakes a different number of hands.

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